\begin{align} \text{area} & = \pi r^2 \\[10pt] \text{circumference} & = 2\pi r \\[10pt] \frac{\text{circumference}}{2\pi} & = r \\[10pt] \left(\frac{\text{circumference}}{2\pi} \right)^2 & = r^2 \\[10pt] \pi \left(\frac{\text{circumference}}{2\pi} \right)^2 & = \pi r^2 = \text{area} \\[10pt] \frac{\text{cicumference}^2}{4\pi} & = \text{area} \end{align}

I think not. Here's why:

$$A = \pi r^2$$ $$C = 2 \pi r$$

Now take

$$A = C^2$$ $$\implies \pi r^2 = (2 \pi r)^2 = 4 \pi^2 r^2$$ $$\implies \pi = 4 \pi^2 $$ $$\implies \frac{1}{4} = \pi $$

A more convincing argument would solve for the radius: $$\pi r^2 = 4 \pi^2 r^2$$ $$\implies \pi r^2 - 4 \pi^2 r^2 = r^2(\pi - 4\pi^2) = 0$$ $$\implies r = 0$$

I will restate my comment: if you think about it, the circumference squared would mean a square where the two sides were the length of the circumference. Since this is necessarily larger than the diameter of the circle, the circle would certainly fit fully inside of the resulting square.

Here is a picture showing that there is no way for the two areas to be equal. The square created from "squaring" the circumference is much much bigger than the original circle (I show 9 circles fitting inside--but there is a lot of unused area--it should be exactly $4\pi$ times the original area of the circle--so there are about 3 and a half more circles that could fit in that area).